If the boiling point of an aqueous solution is `100.1^(@)C`, what is its freezing point ? Given `l_(f)=80` ,` l_(v)=540 "cal" g^(-1)` respectively, of `H_(2)O`.

Explanation

`DeltaT_(b)=K_(b)xx m`
`DeltaT_(f)=K_(f)xx m`
` (DeltaT_(b))/(DeltaT_(f))=K_(b)/K_(f)=(RT_(b)^2)/(1000l_(v))xx(1000l_(f))/(RT_(f)^2)`
`(DeltaT_(b))/(DeltaT_(f))=(T_(b)^(2)xxl_(f))/(T_(f)^(2)xxl_(v))`
`T_(b)=100 + 273 =373 K`
`T_(f)=273 K`
`DeltaT_(b)=100.1-100`
`=0.1^(@)C`
`0.1/(DeltaT_(f))=((373)^(2)xx80)/((273)^(2)xx540)`
`DeltaT_(f)=0.36^(@)C`
`T_(f)=-0.36^(@)C`.